The fixed point for base (eta + eps) is e + delta(eps), where delta satisfies:

Now if is on the order of , we have , so the effect of an additional level of tetration is to add to the exponent. To cross this region of length would require between and steps.

But if is of a larger order, the epsilon-dependent terms may be neglected, and we get that .

So it takes roughly tetration levels to reach .

To be continued...

Now . So we see that is e for any constant from to . Also, it is e for any constant less than 2e, since . Similarly, for any positive sigma.

But assuming positive sigma again, would be exponentiated times and then another ; the first operation is enough to move the exponent up to at least , while the next would take it from there to and each additional step would square the excess over e again. So we would get at least , which clearly tends to infinity as epsilon approaches zero.

So is e if sigma is 0 or positive (probably independent of C), and infinite if sigma is negative.

bo198214 Wrote:Oh, you mean we have an upper fixed point of the tetrational for and the fixed point can then be computed by or . Ya, interesting. I dont know whether we even have a thread on the forum that dealt with the topic of the fixed point of tetrationals.
Of course there maybe always the dependency of the values from the chosen method of tetration.

slog_b (x) = x <=> b ^^ x = x => b ^^^ oo = x
yeah!

to remember:
one - 1^oo = 1
Euler - (e^(1/e)) ^^ oo = e

now new fixed point
(1.63532...) ^^^ oo ~= (3.08855...) what is this new result? it's a Super-Euler?

bo198214 Wrote:So how big is the difference between both methods, with respect to the computed fixed point?

I calculated Nuninho's constant, to 32 decimal digits of precision using my latest kneser.gp program. For bases around 1.6, it takes about 70 seconds to generate sexp accurate to 32 decimal digits. Then we generate the Taylor series, centered around 3.0. Then we generate the Taylor series for sexp'(x), centered around 3.0. Calculate when sexp'(x)=1. Now we have the 'x' value for the local minimum of sexp(x)-x. For that x, calculate sexp(x)-x. If sexp(x)-x>0, then the current base is bigger than Nuinho's constant; if sexp(x)-x<0, then the current base is smaller than Nuinho's constant. Do a binary search .... Here's the result, accurate to 32 decimal digits.

Here is a graph, showing sexp(x), and the line f(x)=x. The two graphs intersect each other at the lower fixed point, and at the upper fixed point. At the lower fixed point, the slope>1, so regular iteration is well defined. The slope at the upper fixed point=1, so this is a parabolic fixed point, much like eta.
- Sheldon

sexp taylor series, centered at the upper fixed point. Parabolic regular iteration, since sexp(upfixed)-upfixed=0, and the derivative=1. However, the pentation series above was developed from the lower fixed point. It might also be interesting to develop the pentation from the upper fixed point.